Question

In the network shown, the equivalent resistance between $P$ and $Q$ is $\frac{4}{3}\Omega$. Hence the value of $r$ is

• A. $3\Omega$
• B. $4\Omega$
• C. $5\Omega$
• D. $6\Omega$

Answer 1

The correct option is D $6\Omega$

R.E.F.Image.

$R_{net}=\frac{4\times \left(\frac{3r}{3+r}\right)}{4+\frac{3r}{3+r}}=\frac{4}{3}$

$\frac{3r}{7r+12}=\frac{1}{3}$

$9r=7r+12$

$r=6\Omega$

Option (D)